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Dãy số có 2 chữ số chia hết cho 3 là:[12,15,....,99]
Khoảng cách của từng số hạng là 3
Số số hạng là: (99-12):3+1=30(số)
Vậy có 30 số có 2 chữ số chia hết cho 3
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{42}{43}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+\frac{42}{43}\)
\(=6-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)+\frac{42}{43}\)
\(=6-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\right)+\frac{42}{43}\)
\(=6-\left(1-\frac{1}{6}\right)+\frac{42}{43}\)
...
bn tự tính tiếp nha
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
(273+485)*16-483:3*4
=758*16-483:3*4
=12128-161*4
=12128-644
=11484
nhớ cick cho mk nhé bạn!
5/2+13/6+25/12+41/20+61/30+85/42
=2+1/2+2+1/6+2+1/12+2+1/20+2+1/30+2+1/42
=12+(1/2+1/6+1/12+1/20+1/30+1/42)
=12+(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7)
=12+(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)
=12+1-1/7
=90/7
5/2+13/6+25/12+41/20+61/30+85/42
=2+1/2+2+1/6+2+1/12+2+1/20+2+1/30+2+1/42
=12+(1/2+1/6+1/12+1/20+1/30+1/42)
=12+(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7)
=12+(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)
=12+1-1/7
=90/7
43 x 27 + 93 x 43 + 57 x 61 + 59 x 57
= 43 x ( 27 + 93 ) + 57 x ( 61 + 59 )
= 43 x 120 + 57 x 120
= 120 x ( 43 + 57 )
= 120 x 100
= 1200
b) 64 x 6 + 81 x 4 + 17 x 6
= 6 x ( 64 + 17 ) + 81 x 4
= 6 x 81 + 81 x 4
= 81 x ( 6 + 4 )
= 81 x 10
= 810
A) =43 x (27+93)+57*(61+59)
=43*120+57*120
=120*(43+57)
=120*100
=12000
Theo đề bài :
Vì số cuối là 3072 nên tổng các số trên là 3072
( 3072 + 1 ) x 3072 : 2 = 4720128
Đáp số: 4720128
~~Hok tốt~~